(3) Since \({V}_B^{\prime }\)  is a commutant, and therefore a von Neumann algebra, one
gets the inclusion \({M}_B\subseteq {V}_B^{\prime }\)  directly from
Proposition REF . Now, let us consider the right von
Neumann algebra \({R}_B\)  generated by the Hilbert algebra
\({L}^{1,2}_{B}\) . For a given \(f\in {L}^{1,2}_{B}\) , the associated operator
\(R_f\in {R}_B\)  acts as \(R_f\psi :=\psi \ast f^{-}\) . Then, an explicit check shows that
\(R_f\)  is exactly the integrated operator
\(\frac{1}{2\pi \ell ^2_B}\int _{\mathbb {R}^2}\,\mathrm {d}y\,f(-y)V_B(y)\)  where the \(V_B(y)\) 's are the dual
magnetic translations. As a consequence \({R}_B\subseteq {V}_B\)  and
\({V}_B^{\prime }\subseteq {R}_B^{\prime }\) . The proof is completed by the equality
\({R}_B^{\prime }={M}_B\)  proved in [1]}.
